Finding unit tangent, normal, and binormal vectors for a given r(t)

Question

For my Calc III class, I need to find $T(t), N(t)$, and $B(t)$ for $t=1, 2$, and $-1$, given $r(t)=\{t,t^2,t^3\}$.

I've got Mathematica, but I've never used it before and I'm not sure how to coerce it into solving this. (My professor told us to use a computer, as it starts getting pretty nasty around $T'(t)$. By hand, $$T(t)=\frac{1}{ \sqrt{(1+4t^2+9t^4})}\{1,2t,3t^2\}$$

I've tried defining r as stated above and using D,Norm, and CrossProduct.

However, I get a bunch of Abs in my outputs (am I missing an assumption?). Additionally, I can't seem to google up how to ask Mathematica to sub in a specific value for $t$, once I get the equations worked out properly.

Answer 1

Mathematica wouldn't be much helpful if one applied only formulae calculated by hand.
Here we demonstrate how to calculate the desired geometric objects with the system having a definition of the curve r[t] :

r[t_] := {t, t^2, t^3}

now we call uT the unit tangent vector to r[t]. Since we'd like it only for real parameters we add an assumption to Simplify that t is a real number. Similarly we can do it for the normal vector vN[t] and the binormal vB[t] :

uT[t_] = Simplify[ r'[t] / Norm[ r'[t] ], t ∈ Reals];
vN[t_] = Simplify[ uT'[t]/ Norm[ uT'[t]], t ∈ Reals];
vB[t_] = Simplify[ Cross[r'[t], r''[t]] / Norm[ Cross[r'[t], r''[t]] ], t ∈ Reals]; 

let's write down the formulae :

{uT[t], vN[t], vB[t]} // Column // TraditionalForm

Edit

Definitions provided above are clearly more useful than only to write them down. They are powerful enough to animate a moving reper along a curve r[t]. Indeed the vectors uT[t], vN[t] and vB[t] are orthogonal and normalized, e.g.

Simplify[ Norm /@ {uT[t], vN[t], vB[t]}, t ∈ Reals]

 {1, 1, 1}

To demonstrate a moving reper we can use ParametricPlot3D and Arrow enclosed in Animate :

Animate[
    Show[ ParametricPlot3D[ {r[t]}, {t, -1.3, 1.3}, PlotStyle -> {Blue, Thick}], 
          Graphics3D[{ {Thick, Darker @ Red, Arrow[{r[s], r[s] + uT[s]}]},
                       {Thick, Darker @ Green, Arrow[{r[s], r[s] + vB[s]}]},
                       {Thick, Darker @ Cyan, Arrow[{r[s], r[s] + vN[s]}]}}], 
          PlotRange -> {{-2, 2}, {-2, 2}, {-2, 2}}, ViewPoint -> {4, 6, 0}, 
          ImageSize -> 600],
          {s, -1, 1}]

Answer 2

Observe that even when the tangent vector $r'(t)$ is not normalized, it is still a linear combination of $T(t)$ and $N(t)$. Thus--operating under the usual assumptions that $r'$ and $r''$ exist and are linearly independent--all we have to do is make an orthonormal frame out of $r'(t)$ and $r''(t)$ (which is very much in the spirit of the entire proceeding). It often helps to do the simplification on the spot, so let's include that too. In the following, the calculation itself is the middle line, enclosed by the simplification code:

frame[r_] := Function[{t}, Evaluate[FullSimplify[
  Append[#, Cross @@ #] &@ Orthogonalize[D[r[t], {t, #}] & /@ {1, 2}, Dot], 
  Assumptions -> t ∈ Reals]]]

(If you read the second line in reverse you can tell exactly what it does: take the first two derivatives, orthogonalize them, and throw in their cross product. Including Dot as an option to Orthogonalize does not change the calculation but makes it much easier for Mathematica to simplify the ensuing expressions.)

This object frame transforms any twice-differentiable 3-vector-valued function into a frame-valued function, which returns a list of three 3-vectors; namely, the unit tangent, normal and binormal $\{T(t), N(t), B(t)\}$. For instance, the example in the question can now be addressed by

frame[{#, #^2, #^3} &][t] // TraditionalForm

The method used in frame generalizes to higher (and lower) dimensions: sequentially orthogonalize the first $n-1$ derivatives (in order) and use the wedge product (a generalization of Cross) to obtain the last element of the frame. The wedge product (which is applied to an ordered list of $n-1$ $n$-vectors) can be implemented as Dot[LeviCivitaTensor[Length[#]+1], Sequence @@ #]&.

Answer 3

Mathematica 10 provides new functionality dealing with curves, (see e.g. the Vector Analysis tutorial) like ArcLength, ArcCurvature and especially FrenetSerretSystem:

FrenetSerretSystem[{ x1, ..., xn}, t]  gives the generalized curvatures 
and Frenet-Serret basis for the parametric curve x[t]
i.e.
it returns {{ k1, ..., k(n-1)}, { e1, ..., en}}, where ki are generalized curvatures
and ei are the Frenet-Serret basis vectors. 

Symbolically it yields

 FrenetSerretSystem[ r[t], t] // TraditionalForm

Since we are interested in the Frenet-Seret basis only, having defined

r[t_] := {t, t^2, t^3}

we can evaluate:

FrenetSerretSystem[ r[t], t] // Last // Simplify // Column // TraditionalForm

With FrenetSerretSystem we don't need to define every base vector separately and we can use it also in animations like in the accepted answer.

Answer 4

This is mainly an answer to your last question, but I think it will help with your other ones. I assume you know that the $T$ function is vector valued, and that is what you want.

To substitute in a specific value of $t$, you probably want replacement rules, specifically the ReplaceAll (/.) construct.

For example, if you had defined your expression $T(t)$ as

bigT={1,2t,3t^2}/Sqrt[1+4 t^2+9 t^4]

(I called it bigT because it is not good practice to name expressions as single capital letters, in case they clash with a built in definition.) Then you would type

 bigT/. t-> 2

Or whatever the desired value might be.

The other way to go would be to define a function using SetDelayed. Then it doesn't actually have to be t as the argument. Notice the underscore (Blank) to denote a pattern.

BigTFunction[x_]:= {1,2x,3x^2}/Sqrt[1+4 x^2+9 x^4]

BigTFunction[2]